See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the molecule: The structure shown is the carbonate ion (CO3^2-), which has one central carbon bonded to three oxygen atoms. One bond is drawn as a C=O double bond (bond I), and the other two are drawn as C-O single bonds with negative charges (bonds II and III). Step 2 - Apply resonance: In the carbonate ion, resonance delocalization distributes the pi electron density equally among all three C-O bonds. There are three equivalent resonance structures, each placing the double bond at a different C-O position. As a result, all three C-O bonds are equivalent in the real (resonance hybrid) structure. Step 3 - Determine bond order: Because of full resonance delocalization, the actual bond order of each C-O bond in the carbonate ion is 4/3 (one sigma bond plus 1/3 pi bond character), making all three bonds identical in length. Step 4 - Conclude bond lengths: Since bonds I, II, and III are all equivalent due to resonance, they all have the same bond length. The drawn Lewis structure showing one double bond and two single bonds is merely a resonance contributor, not the actual structure. Step 5 - Why other options fail: - Option (a) I > II = III: This would be correct only if bond I were longer than II and III, which contradicts resonance equivalence. - Option (b) II > III > I: This implies all bonds are different, ignoring resonance entirely. - Option (c) I > III > II: Same issue as (b), ignoring resonance. - Option (d) II = III = I: Correctly reflects that resonance makes all three C-O bonds equivalent in length. Therefore, the correct answer is D.